Solve 7x=2x^2-4

$$7 x = 2 x ^ { 2 } - 4$$

Solve for x

$$x=4$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$7x-2x^{2}=-4$$

Add 4 to both sides.

$$7x-2x^{2}+4=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}+7x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\times 4=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=8 b=-1$$

Rewrite -2x^{2}+7x+4 as \left(-2x^{2}+8x\right)+\left(-x+4\right).

$$\left(-2x^{2}+8x\right)+\left(-x+4\right)$$

Factor out 2x in -2x^{2}+8x.

$$2x\left(-x+4\right)-x+4$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(2x+1\right)$$

To find equation solutions, solve -x+4=0 and 2x+1=0.

$$x=4 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$7x-2x^{2}=-4$$

Add 4 to both sides.

$$7x-2x^{2}+4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+7x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 7 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-2\right)\times 4}}{2\left(-2\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-2\right)\times 4}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-7±\sqrt{49+8\times 4}}{2\left(-2\right)}$$

Multiply 8 times 4.

$$x=\frac{-7±\sqrt{49+32}}{2\left(-2\right)}$$

Add 49 to 32.

$$x=\frac{-7±\sqrt{81}}{2\left(-2\right)}$$

Take the square root of 81.

$$x=\frac{-7±9}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-7±9}{-4}$$

Now solve the equation x=\frac{-7±9}{-4} when ± is plus. Add -7 to 9.

$$x=\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-7±9}{-4} when ± is minus. Subtract 9 from -7.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=-\frac{1}{2} x=4$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$7x-2x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+7x=-4$$

Divide both sides by -2.

$$\frac{-2x^{2}+7x}{-2}=-\frac{4}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{7}{-2}x=-\frac{4}{-2}$$

Divide 7 by -2.

$$x^{2}-\frac{7}{2}x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x^{2}-\frac{7}{2}x=2$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=2+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=2+\frac{49}{16}$$

Add 2 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{9}{4} x-\frac{7}{4}=-\frac{9}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=4 x=-\frac{1}{2}$$